Physics Topics such as mechanics, thermodynamics, and electromagnetism are fundamental to many other scientific fields.
What is Kepler’s Law and Gravitational Constant?
Introduction
Since ancient times, scientists have been extremely curious to learn more about the stars and planets.
Since the time of Copernicus, it is known that planets move around the sun. But to investigate the cause and nature of this motion, it was necessary to know the exact positions of the planets in the sky at different times. Astrophysicist Tycho Brahe, for many years, observed the positions of planets without a telescope (Galileo discovered the telescope after the death of Brahe) and published a lot of information about this. Kepler analysed the observational findings of Tycho Brahe and arrived at three laws about planetary motion. These laws are known as Kepler’s laws of planetary motion.
Kepler’s Laws
First law: Every planet moves in an elliptical orbit with the sun at one of its foci.
Second law: The line joining the sun and a planet sweeps out equal areas in equal intervals of time, i.e., the areal velocity of a planet is constant.
Third law: The square of the time period of revolution of a planet is directly proportional to the cube of the length of the semi major axis of its elliptical orbit.
The orbits of planets are known as Keplerian orbits while their motions are known as Keplerian motions.
Proof of Kepler’s Second Law
Let us consider a planet of mass m moving in an elliptical orbit with the sun at focus S [Fig.]. Also, let \(\vec{r}\) be the position vector of the planet with respect to the sun and \(\vec{F}\) be the required centripetal force for the planet.
Torque exerted on this planet by this force about the sun,
\(\vec{\tau}\) = \(\vec{r} \times \vec{F}\) = 0 [∵ \(\vec{r}\) and \(\vec{F}\) are oppositely directed]
But \(\vec{\tau}\) = \(\frac{d \vec{L}}{d t}\) [\(\vec{L}\) = anglular momentum of the planet]
∴ \(\frac{d \vec{L}}{d t}\) = 0 or, \(\vec{L}\) = constant
Now, if the planet moves from position P, to P’ in very small time Δt, then the area swept out by the radius vector \(\vec{r}\) is,
Thus, the areal velocity of the planet remains constant, i.e., the radius vector joining the sun and a planet sweeps out equal areas in equal intervals of time.
Kepler’s third law is mathematically discussed in Section 1.5.1.
Newton’s Law of Gravitation
Sir Isaac Newton analysed these laws of planetary motion and formulated the nature of the force of attraction between the sun and the planets and at the same time established a rule for the measurement of this force. This force of attraction is called gravitation, which acts not only between the sun and the planets, but also between any two particles in the universe.
Definition: The force of attraction between any two particles in the universe is called gravitation.
The rule for the measurement of this force of attraction is known as Newton’s law of gravitation or the law of universal gravitation.
Statement: Any two particles in the universe attract each other along a straight line joining them. The magnitude of this force of attraction is directly proportional to the product of the masses of the particles and inversely proportional to the square of the linear distance between them.
Let two particles of masses m1 and m2 be kept at a distance r from each other [Fig.]. If the force of attraction between them is F, then as per Newton’s law of gravitation
F ∝ m1m2, when r remains constant
and F ∝ \(\frac{1}{r^2}\), when m1 and m2 remain unchanged.
Hence, F ∝ \(\frac{m_1 m_2}{r^2}\) or, F = G\(\frac{m_1 m_2}{r^2}\) ……… (1)
G is called the universal gravitational constant or the gravitational constant.
Equation (1) is the mathematical representation of the law of gravitation.
The masses m1 and m2 used in Newton’s law of gravitation are called gravitational masses. On the other hand, the mass referred to in Newton’s second law of motion is called inertial mass. Though the concept of these two different masses is derived from two different laws, experiments and observations show that the two masses are identical physical quantities.
Hence, it is not necessary to mention these two masses separately.
Vector form of the law of gravitation:
Let \(\vec{r}_1\) and \(\vec{r}_2\) be the position vectors of two particles of mass m1 and m2 respectively [Fig.]. So the position vector of m2 with respect to m1 and that of m1 with respect to m2 are respectively,
Let the unit vector along m2 from m1 is \(\hat{r}_{21}\)
Equations (4) and (5) are the vector form of the law of grav-itation.
Gravitational force between two particles is equal and opposite: From equation (2) and (4) we get,
Gravitational Constant
On substitution of m1 = m2 = 1 and r = 1, equation (1) is reduced to G = F.
This defines the universal gravitational constant.
Definition: The mutually-operative force of attraction between two particles of unit mass kept unit distance apart is called the gravitational constant.
Unit: From equation (1),
unit of G \(=\frac{\text { unit of } F \times(\text { unit of } r)^2}{(\text { unit of } m)^2}\)
Hence, the units of G in different systems are as follows:
Dimension and alternative units: From equation (1),
Following this analysis, we can write the alternative units of G in different systems.
Units stated earlier and those mentioned above are obviously identical.
Magnitude; The magnitude of the universal gravitational constant obtained from the results of different experiments is:
G = 6.67 × 10-8dyn ᐧ cm2 ᐧ g-2 = 6.67 × 10-8 CGS unit
= 6.67 × 10-11 N ᐧ m2 ᐧ kg-2 = 6.67 × 10-11 in SI
As per the definition of G, when two objects of mass 1 kg each are kept lm away from each other, they attract each other withaforceof 6.67 × 10-11N. Clearly, this is a force of very low magnitude. Hence, we cannot feel the gravitational force acting between objects used in our daily lives. But, on the other hand, the force of attraction of the earth is considerable as the mass of the earth is quite immense.
Derivation of Newton’s Law of Gravitation from Kepler’s Law
Let a planet of mass m moves around the sun in a circular orbit of radius r. Let M be the mass of sun, and F is the centripetal force exerted by the sun on the planet.
We know that F = mω2; but ω = \(\frac{2 \pi}{T}\) [where ω = angular velocity and T is period of revolution of the planet around the sun]
Then, F = mr\(\left(\frac{2 \pi}{T}\right)^2\) = \(\frac{4 \pi^2 m r}{T^2}\)
According to Kepler’s third law, T2 = Kr3, where K is constant for all planets.
∴ F = \(\frac{4 \pi^2 m r}{K r^3}\) = \(\frac{4 \pi^2 m}{K r^2}\)
Hence, F ∝ \(\frac{m}{r^2}\)
Therefore, the force F, exerted by the sun on the planet is directly proportional to the mass of the planet and inversely proportional to the square of the distance of the planet from the sun. Since the force is mutual, so, the planet will also exert the same force (in magnitude) on the sun in opposite direction. Thus, the force exerted by the planet on the sun will also directly proportional to the mass M of the sun.
Therefore, F ∝ \(\frac{m M}{r^2}\)
This is the Newton’s law of gravitation.
Universality of the Law of Gravitation
As per Newton’s law of gravitation, the magnitude of the force of attraction between two bodies depends on their masses and the distance between them. This force of attraction does not depend on
- the state of the objects (solid, liquid, gas),
- their chemical composition,
- temperature,
- the intermediate medium, or any other factor.
The law of gravitation is equally applicable for earthly objects separated by small distances and celestial bodies in space, widely separated from one another. It has been possible to explain the motion of planets around the sun satisfactorily using this law. For these reasons, the law of gravitation is taken as a universal law and the gravitational constant is called the universal gravitational constant.
But as per the theory of relativity propounded by Einstein,
- the mass of a body depends on its velocity,
- the distance measured between two objects depends on the velocity of the observer, i.e., the distance between two objects measured by an observer at rest will not be the same as that measured by an observer in motion,
- Newton’s law of gravitation is not applicable in the case of interatomic distances (10-9m).
Because of the reasons stated above, Newton’s law of gravitation cannot be called a universal law any more. But the universality of G has not been denied even in Einstein’s theory.
Numerical Examples
Example 1.
If the distance between the centre of a gold sphere of radius 5 mm and of a lead sphere of radius 11.5 cm is 15cm and the force of gravitational attraction between them is 2.16 × 10-4 dyn, find the value of G, the universal gravitational constant.
Densities of gold and lead are 19.3 g ᐧ cm-3 and 11.3 g ᐧ cm-3 respectively.
Solution:
Mass of gold sphere,
m1 = \(\frac{4}{3}\) × π × (0.5)3 × 19.3 = 10.1 g
Mass of lead sphere,
m2 = \(\frac{4}{3}\) × π × (11.5)3 × 11.3 = 7.2 × 104 g
Given, r = 15 cm
∴ G = \(\frac{F r^2}{m_1 m_2}\) = \(\frac{2.16 \times 10^{-4} \times(15)^2}{10.1 \times 7.2 \times 10^4}\)
= 6.68 × 10-8 CGS unit
Example 2.
Three particles of the same mass are kept at the vertices of an equilateral triangle. The mass of each particle is m and the length of an arm of the triangle is l. Due to the mutual gravitational force of attraction, the particles revolve along the circumcircle of the triangle. Find the velocity of each particle.
Solution:
ABC is an equilateral triangle. At its vertices, three particles A, B and C, each of mass m, are kept [Fig.].
The point of intersection of the medians of ABC is O, which is also the centre of the circumcircle of the triangle. Hence, radius of the circumcircle, r = OA = \(\frac{2}{3}\)AD = \(\frac{2}{3}\)ABsin60° = \(\frac{2}{3} \cdot l \cdot \frac{\sqrt{3}}{2}\) = \(\frac{l}{\sqrt{3}}\).
Hence, the centripetal force needed by each particle to revolve along the circumcircle with velocity v is
F1 = \(\frac{m v^2}{r}\) = \(\frac{m v^2}{\frac{l}{\sqrt{3}}}\) = \(\frac{\sqrt{3} m v^2}{l}\)
Now the gravitational force acting on the particle at point A can be calculated. The force of gravitation on the particle at A due to the particle at B along AB,
F = \(\frac{G \cdot m \cdot m}{(A B)^2}\) = \(\frac{G m^2}{l^2}\)
Component of this force along AE = F cos 60° and that along AD = F sin 60°
Again, gravitational attraction on the particle at A due to the particle kept at C is F along AC.
Component of this force along AH = F cos60° and that along AD = F sin 60°.
Clearly, components along AH and AE cancel each other. Hence, the resultant force F2 on the particle kept at A = sum of the components along AD.
F2 = Fsin60° + F sin60° = 2F sin60°
= 2Fᐧ\(\frac{\sqrt{3}}{2}\) = \(\sqrt{3}\)F = \(\frac{\sqrt{3} G m^2}{l^2}\)
As per the question, this force of attraction due to gravitation supplies the necessary centripetal force. Hence,
F1 = F2 or, \(\frac{\sqrt{3} m v^2}{l}\) = \(\frac{\sqrt{3} G m^2}{l^2}\) or, v2 = \(\frac{G m}{l}\)
or, v = \(\sqrt{\frac{G m}{l}}\)
Example 3.
A mass M is broken into two parts of masses m1 and m2. How are m1 and m2 related so that force of gravitational attraction between the two parts is maximum?
Solution:
Let m1 = m, then m2 = M – m
Gravitational force between the two parts when they are placed at a distance r apart is,
F = G\(\frac{m(M-m)}{r^2}\) = \(\frac{G m M}{r^2}\) – \(\frac{G m^2}{r^2}\)
Differentiating with respect to m, we have,
\(\frac{d F}{d m}\) = \(\frac{G M}{r^2}\) – \(\frac{2 G m}{r^2}\)
For F to be maximum, \(\frac{d F}{d m}\) = 0
∴ \(\frac{G M}{r^2}\) = \(\frac{2 G M}{r^2}\) or, M = 2m or, m = \(\frac{M}{2}\)
∴ m1 = m2 = \(\frac{M}{2}\)
Example 4.
Assuming the earth’s orbit around the sun to be circular, show that the area swept by its radius vector in unit time (areal velocity of the earth) is a constant.
Solution:
Suppose the earth moves in a circular orbit of radius r with the sun at the centre (O) of the orbit [Fig.].
Let the radius vector OP, in an infinitesimal interval of time dt, describe an angle dθ at the centre. Hence, arc PQ = rdθ. As
value of PQ is very small, the arc PQ can be taken to be a straight line (chord PQ).
∴ Area swept in time dt = area of triangle OPQ
= \(\frac{1}{2} O P\) × PQ = \(\frac{1}{2} r \cdot r d \theta\) = \(\frac{1}{2} r^2 d \theta\)
∴ Area swept per unit time = \(\frac{1}{2} r^2 \cdot \frac{d \theta}{d t}\) = \(\frac{1}{2} r^2 \omega\) = constant
[as orbit is circular, r and ω of the earth should be constants]
∴ Area swept in unit time by the radius vector of the earth is a constant.
Example 5.
How fast (in m2/s) is area swept out by
(a) the radius from sun to earth?
(b) the radius from earth to moon? Given distance of sun to earth = 1.496 × 1011 m distance of earth to moon = 3.845 × 108 m and period of revolution of moon = 27\(\frac{1}{3}\) days.
Solution:
(a) The rate, at which the area is swept out by the radius from the sun to the earth,
\(\frac{d A}{d t}\) = \(\frac{\pi r^2}{T}\)
Here, r = 1.496 × 1011 m,
T = 365 days = 365 × 24 × 60 × 60 s
∴ \(\frac{d A}{d t}\) = \(\frac{\pi \times\left(1.496 \times 10^{11}\right)^2}{365 \times 24 \times 60 \times 60}\) = 2.23 × 1015 m2 ᐧ s-1
(b) Here, r = 3.845 × 108m,
T = 27\(\frac{1}{3}\) days = \(\frac{82}{3}\) × 60 × 60 × 24 s
∴ The rate, at which the area is swept out by radius from the earth to moon,
\(\frac{d A}{d t}\) = \(\frac{\pi \times\left(3.845 \times 10^8\right)^2}{\frac{82}{3} \times 24 \times 60 \times 60}\)
= 1.97 × 1011 m2 ᐧ s-1